Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.32 years and the standard deviation is 10.66 years.​ a) Construct a 99% confidence interval for the mean age of all​ customers, assuming that the assumptions and conditions for the confidence interval have been met. ​b) How large is the margin of​ error? ​c) How would the confidence interval change if you had assumed that the standard deviation was known to be 11.0 ​years?

Answer 1

Solution:
Given in the question
Total no. of samples = 25
Mean. =32.32
Standard deviation = 10.66
alpha = 0.01 and alpha/2 = 0.005
Here sample size is 25 and less than 30 so we will calcuate talpha/2 = 2.58 at alpha/2 = 0.005 and df = (25-1)=24
so confidence can be calculate as
Mean +/- Talpha/2 * Standard deviation/sqrt(n)
32.32 +/- 2.58*10.66/sqrt(25)
32.32+/-2.58*10.66/5
32.32 +/- 2.58*2.13
32.32 +/- 5.4917
So confidence interval is
26.8283 to 37.8117

Solution(b)
Margin of error = talpha/2*standard deviation /sqrt(n) = 5.4917

Solution(c)
If Standard deviation = 11
than confidence interval is
32.32 +/- 2.58*11/sqrt(25) = 32.32 +/- 2.58*2.2 = 32.32 +/- 5.666
So confidence interval is
26.6532 to 37.9868
As we increase standard deviation, margin of error increased from 5.4917 to 5.666 and confidence interval also increased.
Confidence increase from (26.8283, 37.8117) at 10.66 standard deviation to (26.6532,37.9868) at 11 standard deviation.